4.4 Axioms for the reals and first approximations

Why define the reals before constructing them?

In earlier chapters, the pattern was:

1. decide what structure we want,
2. build a model that has that structure,
3. check the construction carefully.

For the real numbers, the target is a complete ordered field. Specifying that target tells us exactly what a construction must establish.

That is good mathematical practice. If you do not know the target properties first, then even a beautiful construction can feel unmotivated.

The formal target

Definition

Model of the real numbers

A set $R$ is a model of the real numbers if it is equipped with:

• elements $0,1\in R$,
• binary operations $+$ and $\cdot$,
• a total order $\le$,

such that:

• $(R,0,1,+,\cdot,\le)$ is an ordered field;
• $(R,\le)$ is complete.

This definition compresses the whole chapter into one sentence:

the reals are an ordered field with no missing least upper bounds or greatest lower bounds.

Theorem

Uniqueness of the real numbers

Any two complete ordered fields are isomorphic as ordered fields. Thus different valid constructions can be treated as models of the same $R$.

The uniqueness theorem is stated here without proof. It explains why we can move between different constructions of $R$. Dedekind cuts and Cauchy sequences look different internally, but once both constructions satisfy the complete ordered field axioms, they represent the same real number system up to the unique structure that matters here: addition, multiplication, and order.

Worked example

Why Q is not yet a model of the reals

The rational numbers already satisfy the ordered-field part. What fails is completeness.

The set

$$S=\{x\in Q\mid x^2\lt 2\}$$

is nonempty and bounded above in $Q$, but it has no rational supremum. So $Q$ is close to the target, yet still not the real-number system.

Why decimal expansions are not the full story

A first instinct is to say:

"A real number is just an infinite decimal expansion."

That instinct is useful, but it is not yet the cleanest definition.

There are at least two reasons.

• The same real number can have more than one decimal expansion, as in $0.9999\ldots = 1$.
• Decimal expansions force a specific base, such as base 10, even though the idea of the real numbers should not depend on that arbitrary choice.

So we should not reject decimal intuition. Instead, use it as a guide toward a more structural construction.

The approximation idea

Suppose a real number is written informally as

$$r=10.234890234809234\ldots$$

Then we can approximate $r$ from below and above by rational numbers:

$$10\lt r\lt 11,$$ $$10.2\lt r\lt 10.3,$$ $$10.23\lt r\lt 10.24,$$

and so on.

Each additional digit narrows a rational interval. The eventual construction must explain why all these finite comparisons determine one boundary.

Worked example

Lower and upper rational fences

For a decimal expansion such as $r=3.1415\ldots$, the first few rational fences look like

$$3\lt r\lt 4,\qquad 3.1\lt r\lt 3.2,\qquad 3.14\lt r\lt 3.15.$$

Each extra digit gives a narrower rational interval containing $r$.

Dividing the rationals into two camps

From this approximation viewpoint, the real number $r$ determines two sets of rationals:

$$A=\{q\in Q\mid q\lt r\}, \qquad B=\{q\in Q\mid q\ge r\}.$$

The important idea is not the notation itself. The important idea is that a real number can be read as a boundary that separates the rationals into a left camp and a right camp.

That is exactly the motivation behind the Dedekind-cut construction in the next part of the construction. This stage is still motivational rather than fully formal, but it already tells you what kind of object a real number should be: something that organizes the rationals by comparison.

From an approximation picture to a construction

The sets just displayed use a real number $r$ to describe which rationals lie below it. They motivate a construction but do not yet constitute one: a construction must define the boundary without assuming that $r$ already exists. The next note does this with axioms for a rational lower set. Before that step, we establish what the complete ordered-field target implies about approximation.

The Archimedean consequence without circularity

The approximation picture must not quietly assume the Archimedean property as an unexplained fact about $R$. In the complete ordered-field target, it is derived from completeness. Suppose, for contradiction, that the natural numbers were bounded above in $R$. Completeness would give

$$s=\sup\{0,1,2,\ldots\}.$$

Because $s-1$ is smaller than $s$, it cannot be an upper bound. Hence some natural number $n$ satisfies $s-1\lt n$, and then $n+1\gt s$, contradicting that $s$ was an upper bound. Therefore the natural numbers are unbounded above. For every real $x$ there is a natural $n$ with $n\gt x$; applying this to $1/\varepsilon$ gives the usual small-mesh estimates.

Theorem

Archimedean approximation principle

For every $x\in R$ there is $n\in N$ with $n\gt x$. Equivalently, for every $\varepsilon\gt 0$ there is $n\in N$, $n\ge1$, such that $1/n\lt \varepsilon$. The proof uses completeness and does not assume that a decimal expansion has already constructed the real line.

Worked example

Choosing a rational fence width

Given $\varepsilon\gt 0$, choose $n\gt 1/\varepsilon$. Then $1/n\lt \varepsilon$. If an approximation interval has width at most $1/n$, its width is therefore less than $\varepsilon$. This is the quantitative reason that successive decimal or rational fences can be made arbitrarily narrow.

Common mistake

Do not use the conclusion as the construction

Saying “choose a decimal expansion of the real number” before proving that the fences converge assumes the very boundary that the construction is meant to produce. First establish the order and completeness principles; then use the Archimedean consequence to control the approximation width.

Density and rational fences

The Archimedean property does more than assert abstract unboundedness: it lets us place rational numbers around any real number with a prescribed accuracy. The complete mesh argument is written explicitly in the next section. It uses a least-integer choice and then divides by a positive integer; the resulting rational fence works in every base and does not depend on decimal notation.

Theorem

Rational density with a requested error

For every $x\in R$ and every $\varepsilon\gt 0$, there is $q\in Q$ such that $|x-q|\lt \varepsilon$. Apply the interval argument below to $x-\varepsilon$ and $x+\varepsilon$.

Worked example

A rational inside a specified interval

Choose $n=1000$, so $1/n=0.001\lt 1.42-1.41$. The least integer strictly above $1000(1.41)=1410$ is $m=1411$. Hence $m/n=1.411$ lies strictly inside $(1.41,1.42)$. Choosing $n=100$ would give the right endpoint $1.42$, so the strict mesh inequality is essential.

Worked example

Why completeness is still needed

Density says that rational points occur between any two real points. It does not say that a bounded rational subset has a rational endpoint. The set below $\sqrt{2}$ can be densely populated by rationals while its boundary remains an irrational real. Approximation and attainment are different claims.

Trichotomy, absolute value, and error control

The total-order axiom gives a three-way decision for every pair of real numbers. Exactly one of $x\lt y$, $x=y$, or $y\lt x$ holds. In particular, every real number is positive, zero, or negative, and the sign of a difference determines which side of an interval it occupies. The three alternatives are exhaustive and mutually exclusive, so later arguments may split into cases without leaving an unclassified possibility.

Theorem

Triangle inequality for rational error terms

For $u,v\in Q$ (and, once the ordered-field axioms are established, for $u,v\in R$),

$$|u+v|\le |u|+|v|.$$

Consequently, if $|x-q|\lt\varepsilon$ and $|q-r|\lt\delta$, then $|x-r|\lt\varepsilon+\delta$.

To prove the first statement, if $u+v\ge0$, then $|u+v|=u+v\le|u|+|v|$ because $u\le|u|$ and $v\le|v|$. If $u+v\lt0$, then $|u+v|=-(u+v)=(-u)+(-v)\le|u|+|v|$, since $-u\le|u|$ and $-v\le|v|$. The two sign cases cover every possibility. For the consequence, write $x-r=(x-q)+(q-r)$ and apply the inequality:

$$|x-r|\le|x-q|+|q-r|\lt\varepsilon+\delta.$$

This is why a mesh estimate is useful: errors add in a controlled way rather than merely becoming “visually small.”

The rational mesh written without a hidden choice

The density argument can be made completely explicit. Given $x\lt y$ in $R$, first choose $n\in N$, $n\ge1$, with $1/n\lt y-x$. By the Archimedean property, the set of integers strictly greater than $nx$ is nonempty. It has a least element $m$ by the well-ordering of $N$ after translating by a sufficiently large integer. Minimality gives

$$m-1\le nx\lt m.$$

Since $n>0$, division preserves the inequalities. Thus

$$x\lt\frac mn\le x+\frac1n\lt y.$$

The number $m/n$ is rational and lies strictly in the requested interval. The strict inequality $1/n\lt y-x$ is essential: if one used merely $1/n\le y-x$, the constructed fraction could land exactly at $y$.

Worked example

A three-part approximation question

Suppose $x\in R$ and $\varepsilon>0$. Choose $n>2/\varepsilon$ and apply the mesh construction to the interval $(x-\varepsilon/2,x+\varepsilon/2)$. The resulting rational $q$ satisfies $|x-q|\lt\varepsilon/2\lt\varepsilon$. If a second rational $r$ satisfies $|q-r|\lt\varepsilon/2$, the triangle inequality gives $|x-r|\lt\varepsilon$. Thus the same proof simultaneously produces a rational approximation and a rigorous tolerance for replacing one approximation by another.

Worked example

A trichotomy check around a rational fence

Suppose a real $x$ satisfies $1.41\lt x\lt1.42$. Trichotomy says that exactly one of $x\lt1.411$, $x=1.411$, or $1.411\lt x$ holds. If the third case holds, then $q=1.411=1411/1000$ is a rational strictly below $x$ and still inside the original interval. If the first case holds, use $1.41$ as the lower fence and repeat with a finer mesh. The equality case is not an error: it means the rational fence has already reached $x$. Keeping all three cases prevents a strict inequality from being silently substituted for a non-strict one.

The roles of the assumptions are now visible. Completeness supplies the target real boundary and, through the supremum argument, the Archimedean property. Trichotomy allows a definite side comparison. Well-ordering chooses the least integer needed by the mesh. Field order compatibility permits division by the positive integer $n$. Rational density is the result of these ingredients together; it is not an additional replacement for completeness.

There is a useful distinction between an approximation procedure and a construction of a number. Once a real $x$ is already available, the mesh lemma can locate rationals around it and the triangle inequality can compare their errors. A construction must go further: it must specify which approximation data count as the same object, prove that the resulting object supports the field operations, and show that the order is complete. Decimal notation alone does not perform those tasks. The left/right split of the rationals is motivating because it records the boundary data, while the Dedekind-cut note will make that data into a formal object.

The order of dependence matters. The mesh width is chosen only after the Archimedean estimate has been established. The Archimedean estimate is derived from completeness by applying the least-upper-bound principle to the natural numbers, so it cannot be used as a hidden premise in that derivation. Once the estimate is available, rational density follows from well-ordering and positive division. This dependency chain keeps the first approximation argument noncircular and makes clear which part belongs to the axiomatic target and which part belongs to the later construction.

Quick checks

Checkpoint

Which property does Q fail, so that it cannot be a model of the real numbers?

Compare Q with the formal definition above.

Solution · Answer

$Q$ fails completeness. It is an ordered field, but not every nonempty bounded subset of $Q$ has its supremum and infimum inside $Q$.

Checkpoint

What information do the inequalities $10.23<r<10.24$ give you about r?

State it in terms of rational approximation.

Solution · Answer

They place $r$ inside a narrow rational interval. In other words, they give a lower rational approximation and an upper rational approximation that trap $r$.

Exercises

Checkpoint

Let $x\in R$ and $\varepsilon>0$. Find rational approximations $q,r$ so that $|x-q|<\varepsilon/3$ and $|q-r|<2\varepsilon/3$, then prove $|x-r|<\varepsilon$.

Apply rational density twice and keep track of the two error budgets.

Solution · Model solution

Rational density gives $q\in Q$ with $|x-q|<\varepsilon/3$. Apply it again to $q$ with tolerance $2\varepsilon/3$ to obtain $r\in Q$. The triangle inequality then gives

$$|x-r|\le|x-q|+|q-r|<\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon.$$

The tolerances can be unequal; what matters is that their sum stays within the requested error budget. No particular decimal expansion is needed.

Checkpoint

Suppose we try to construct real numbers from infinite digit strings. Besides assigning a boundary to each string, what equality and arithmetic checks must the construction establish?

Use $1.25000\ldots=1.24999\ldots$ to test the meaning of equality.

Solution · Model solution

The construction must identify different strings that describe the same boundary. Addition, multiplication, and order must be independent of which representative string is chosen. Finally it must verify the ordered-field laws and completeness. An approximation notation is useful, but does not establish these structural properties on its own.

Related notes

Read this after 4.3 Completeness and gaps in Q. Then continue with 4.5 Dedekind cuts and the embedding of Q, where the informal left/right split of $Q$ becomes the first fully rigorous construction of the real numbers.